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Suppose that $\phi : G \rightarrow H$ is a lie group homomorphism. How do I show that $\ker(\phi)$ is a closed subgroup of $G$? In general, when we refer to $H \leq G$ being closed, do we mean that $H$ is closed under the topology associated with $G$? Assuming that $\phi$ is a continuous map, we have that $\phi^{-1}(1) = \ker(\phi)$, since the inverse of a closed set is closed, we have that $\ker(\phi)$ is closed. Is this correct reasoning?

user100101212
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    Yes, your reasoning is correct. – Albert Feb 29 '20 at 19:24
  • Is it true that if $H \leq G$ is a closed lie subgroup, then it is an embedded lie subgroup? – user100101212 Feb 29 '20 at 19:27
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    Yes, actually something stronger is true: If $H$ is an abstract subgroup of $G$ (i.e. a subgroup but not necesarilly a Lie subgroup) and $H$ is a closed subset of $G$, then $H$ is an embedded Lie subgroup of $G$. It is a highly non trivial theorem known as the closed subgroup theorem. You may want to take a look to Warner's book or to Knapp's book on Lie groups. – Albert Feb 29 '20 at 19:33

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